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21 Jan 2010, 17:24
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Working independently at their respective constant rates, pumps X and Y took 48 minutes to fill an empty tank with water. What fraction of the water in the full tank came from pump X ?
(1) Working alone at its constant rate, pump X would have taken 80 minutes to fill the tank with water.
(2) Working alone at its constant rate, pump Y would have taken 120 minutes to fill the tank with water.
(A) Statement 1 alone is sufficient, but statement 2 alone is not.
(B) Statement 2 alone is sufficient, but statement 1 alone is not.
(C) Both statements together are sufficient, but neither alone is sufficient.
(D) Each statement alone is sufficient.
(E) Neither statement is sufficient, even when combined.
(1) Working alone at its constant rate, pump X would have taken 80 minutes to fill the tank with water.
(2) Working alone at its constant rate, pump Y would have taken 120 minutes to fill the tank with water.
(A) Statement 1 alone is sufficient, but statement 2 alone is not.
(B) Statement 2 alone is sufficient, but statement 1 alone is not.
(C) Both statements together are sufficient, but neither alone is sufficient.
(D) Each statement alone is sufficient.
(E) Neither statement is sufficient, even when combined.
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23 Jul 2010, 14:44
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1. Working independently at their respective rate, pump X and Y took 48 minutes to fill an empty tank with water. what fraction of the water in the full tank came from pump X?
Let the rate of pump X be \(x\) liters per minute, the rate of pump Y be \(y\) liters per minute and the capacity of the tank be \(c\) liters.
Given: \((x+y)*48=c\). Question: \(\frac{48x}{c}=?\) (As from the pump X in 48 minutes would come \(48x\) liters)
(1) Working along at its constant rate, pump X would have taken 80 minutes to fill the tank with water --> \(80x=c\) --> \(\frac{48x}{c}=\frac{48x}{80x}=\frac{3}{5}\). Sufficient.
(2) Working alone at its constant rate, pump Y would have taken 120 minutes to fill the tank --> \(120y=c\) --> as \((x+y)*48=c\), then \((x+y)*48=120y\) and \(48x=72y\) --> \(\frac{48x}{c}=\frac{72y}{120y}=\frac{3}{5}\). Sufficient.
Answer: D.
Hope it's clear.
Let the rate of pump X be \(x\) liters per minute, the rate of pump Y be \(y\) liters per minute and the capacity of the tank be \(c\) liters.
Given: \((x+y)*48=c\). Question: \(\frac{48x}{c}=?\) (As from the pump X in 48 minutes would come \(48x\) liters)
(1) Working along at its constant rate, pump X would have taken 80 minutes to fill the tank with water --> \(80x=c\) --> \(\frac{48x}{c}=\frac{48x}{80x}=\frac{3}{5}\). Sufficient.
(2) Working alone at its constant rate, pump Y would have taken 120 minutes to fill the tank --> \(120y=c\) --> as \((x+y)*48=c\), then \((x+y)*48=120y\) and \(48x=72y\) --> \(\frac{48x}{c}=\frac{72y}{120y}=\frac{3}{5}\). Sufficient.
Answer: D.
Hope it's clear.
25 Jan 2011, 20:18
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luminousmocha
You can look at it in another way
Stmnt 1: Time taken by X alone : Time taken by X and Y = 80:48 = 5:3
Rate of work of X : Rate of work of X and Y = 3:5 (Work done in both cases is the same. If X alone takes more time, its rate of work is smaller)
So X does 3/5 of total work. Sufficient.
Stmnt 2: Same calculations for y alone with give you that Y does 2/5 of the total work. Sufficient.
Answer D
